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So I have a continuous distribution with CDF $F(x) = 1 - \exp((-x^2)/2), x>0$.

For the mean, I got the derivative $f(x) = x\exp((-x^2)/2)$, but the integral for $E(X)$ gives me $-\sqrt{2\pi}/2$, which must not be right, since the CDF is only defined for $x>0$.

Also, any tips on how to calculate the variance, because I feel the formula $E(X^2) - \mu^2$ will lead me to similarly strange results. Is this the CDF of a known distribution that I am missing?

MPW
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1 Answers1

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Note that $X> 0$ with probability one. Then we can use the fact that in this case $$ EX=\int_0^\infty 1-F(x)\, dx=\int_{0}^{\infty}e^{-x^2/2}\, dx=\frac{\sqrt{2\pi}}{2}\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}e^{-x^2/2}=\frac{\sqrt{2\pi}}{2}. $$ Next $$ EX^2=\int_0^\infty2x(1-F(x))=\int_0^\infty 2x(e^{-x^2/2})\, dx=\int_0^\infty e^{-u/2}\, du=2. $$

Sri-Amirthan Theivendran
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  • I actually found the mistake in my computation for the mean, and I got the correct result. For the variance, how did you derive this formula? Is it standard? Because i only knew the formula with the integral of $((x - μ)^2)f(x)$ –  Feb 22 '19 at 00:34
  • @JBuck See this answer – angryavian Feb 22 '19 at 07:23